The minimum stopping distance when the car is moving at 32.0 m/s is 348.3 m.
<h3>
Acceleration of the car </h3>
The acceleration of the car before stopping at the given distance is calculated as follows;
v² = u² + 2as
when the car stops, v = 0
0 = u² + 2as
0 = 15² + 2(76.5)a
0 = 225 + 153a
-a = 225/153
a = - 1.47 m/s²
<h3>Distance traveled when the speed is 32 m/s</h3>
If the same force is applied, then acceleration is constant.
v² = u² + 2as
0 = 32² + 2(-1.47)s
2.94s = 1024
s = 348.3 m
Learn more about distance here: brainly.com/question/4931057
#SPJ1
Answer:
Measure the brightness of a star through two filters and compare the ratio of red to blue light. Compare to the spectra of computer models of stellar spectra of different temperature and develop an accurate color-temperature relation.
Answer:
mass = 61.16 kg
Explanation:
given data
distance = 50 m
energy = 500 W = 500 × 60sec = 30000 J/min
to find out
mass
solution
we will apply here energy equation
that is
energy = m×g×h .............1
put here all value m mass here and g = 9.81 and h is distance
energy = m×g×h
30000 = m×9.81×50
mass = 61.16 kg
cardiovascular fitness: 3, 4, 7
flexibility: 1, 5
muscular fitness: 2, 6